8 new employees are assigned to 3 different departments. How many differents ways of assigning the employees exist if each department has to receive at least one employee?
At first I thought to work with the departments as sets but I think that is not possible beacuse the problem does not say that the employees can work in one or more departmets. Is working with the departments as sets the right option?
The number of ways to map $8$ people into $3$ departments is $N(0)=3^8$. The number of ways to miss a particular department would be $\binom{3}{2}2^8$. The number of ways to miss two particular departments would be $\binom{3}{1}1^8$. The number of ways to miss three particular departments would be $\binom{3}{0}0^8$.
Inclusion-Exclusion says there are $$ 3^8-\binom{3}{2}2^8+\binom{3}{1}1^8-\binom{3}{0}0^8=5796 $$ ways to distribute $8$ people among $3$ departments with no empty departments.
This is equivalent to count the number of surjective maps from the set of employees $\{e_1,\dots,e_8\}$ to the set of departments $\{d_1,d_2,d_3\}$. So you get $$S(8,3)\times 3!=5796\quad \text{ways}$$ where $S(8,3)$ is the Stirling number of second kind. Recall that the condition each department has to receive at least one employee means that each map has to be surjective.
The number of surjective maps from an $n$ elements set onto a $k$ elements set is given by: $$S(n,k)\cdot k!$$
